Problem 35 A circular ring of wire of radiu... [FREE SOLUTION]

91影视

Calculus

Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason

$Math Studyset 91影视 Explanations$ Math

6 Edition

Chapter 4: Problem 35

A circular ring of wire of radius $r_{0}$ lies in a plane perpendicular to the $x$ -axis and is centered at the origin. The ring has a positive electric charge spread uniformly over it. The electric field in the $x$ -direction, $E,$ at the point $x$ on the axis is given by $$E=\frac{k x}{\left(x^{2}+r_{0}^{2}\right)^{3 / 2}} \quad \text { for } \quad k>0.$$ At what point on the $x$ -axis is the field greatest? Least?

Short Answer

Expert verified

The field is greatest at $x = 0$ and least as $x \to \pm \infty$.

Step by step solution

Understand the Problem

We are given an electric field function $E$ for a charged ring, dependent on $x$ and a constant radius $r_0$. We need to determine where this electric field is strongest and weakest along the $x$-axis.

Differentiate the Function

To find where the field is greatest (maximum) or least (minimum), we take the derivative of $E$ with respect to $x$. We apply the quotient rule to the function $E = \frac{kx}{(x^2 + r_0^2)^{3/2}}$.

Apply the Quotient Rule

The quotient rule for differentiating $ \frac{u}{v}$ is $ \frac{u'v - uv'}{v^2}$. Here, $u = kx$ and $v = (x^2 + r_0^2)^{3/2}$. Calculate $u' = k$ and $v'$ using the chain rule, which gives $v' = \frac{3}{2}(x^2 + r_0^2)^{1/2}(2x)$.

Simplify the Derivative

Substitute $u, v, u', \text{and} \; v'$ into the quotient rule formula. Simplify to get the derivative $E'(x)$. This involves canceling terms and factoring where possible.

Set Derivative to Zero and Solve

Solve $E'(x) = 0$ to find the critical points. Set the numerator of the derivative equal to zero and solve for $x$. This may involve simplifying expressions and possibly using algebraic techniques or approximations.

Analyze the Critical Points

Evaluate $E(x)$ at critical points and at the boundaries. Since this is a physical scenario, $x$ extends from $-\infty$ to $+\infty$. Consider behavior as $x \to \pm \infty$ and near $x = 0$.

Find Maximum and Minimum

Determine which of the critical points gives the maximum and which gives the minimum value for $E(x)$. Compare values computed in the previous step to conclude.

Conclusion

Summarize the results of when $E(x)$ is at maximum and minimum. Ensure you mention the values of $x$ where these occur.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule

When dealing with complex functions involving division, like our electric field function, the quotient rule is a lifesaver. It helps us differentiate expressions where one function is divided by another. Given a function $ E = \frac{u}{v} $, the quotient rule states that the derivative $ E'(x) $ is determined by the formula:\[ E'(x) = \frac{u'v - uv'}{v^2} \]Here's how it works:

Identify $ u $ and $ v $ in your function. For our exercise, $ u = kx $ and $ v = (x^2 + r_0^2)^{3/2} $.
Find the derivative of the numerator, $ u' = k $, since the derivative of a linear function is a constant.
Calculate the derivative of the denominator. Here, $ v $ requires us to use the chain rule for differentiation, resulting in $ v' = \frac{3}{2}(x^2 + r_0^2)^{1/2}(2x) $.
Substitute $ u, v, u', \text{and}\ v' $ into the quotient rule formula to find $ E'(x) $.

Understanding and applying the quotient rule is essential for solving problems involving electric fields and other physics equations that feature complex rational functions.

Critical Points

Critical points are essential to analyze because they indicate where a function's value might be at a maximum, minimum, or a flat point (neither). To find these points for our electric field function $ E $, we solve $ E'(x) = 0 $ (where the derivative is zero). This gives us potential points of change in the electric field strength along the $ x $-axis.Here's the breakdown:

Derive $ E(x) $ using the steps from the quotient rule. This will give us $ E'(x) $, the slope of the electric field with respect to $ x $.
Set $ E'(x) = 0 $. A zero derivative means there's no slope at that point鈥攊ndicating a critical point.
Solving $ E'(x) = 0 $ usually involves some algebraic manipulation to find the values of $ x $ where this condition holds.

Critical points are crucial when optimizing fields, as they help identify spots where the electric field changes direction or intensity.

Derivative Analysis

Analyzing the derivative of the electric field function provides insight into how the field behaves over different regions on the $ x $-axis. By analyzing $ E'(x) $, we can determine whether the electric field is increasing or decreasing at any point.Understand these steps for derivative analysis:

Use $ E'(x) $ to investigate how the electric field changes relative to changes in $ x $.
Check the sign of $ E'(x) $:
- If $ E'(x) > 0 $, the field is increasing at that point.
- If $ E'(x) < 0 $, the field is decreasing at that point.
Consider both critical points and end behavior (as $ x \to -\infty $ and $ x \to +\infty $) to understand overall trends.

By combining these observations, derivative analysis allows us to draw a detailed picture of the maximum and minimum values of the electric field and effectively optimize it for the given physical scenario.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.